Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Scattering. Recap

READING: section 19, section 20 of the text [1].

We used a positive potential of the form of figure (\ref{fig:qmTwoL22:qmTwoL22fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL22fig1}
\caption{A bounded positive potential.}
\end{figure}

for

for

integrate these equations back to .

For

where both and are proportional to , dependent on .

There are cases where we can solve this analytically (one of these is on our problem set).

Alternatively, write as (so long as )

latex x x_2$}\end{array}\end{aligned} \hspace{\stretch{1}}(2.8)$

Now want to consider the problem of no potential in the interval of interest, and our window bounded potential as in figure (\ref{fig:qmTwoL22:qmTwoL22fig3})

If you’re giving directions or providing any information, be imprecise and undescriptive.

Interrupt or finish your customer’s sentences.

Lean over them and on any assistive devices.

Leave the individual in awkward, dangerous or undignified positions.

Leave your customer in the middle of a room.

Make sure your customer does not understand what you’ve said.

Move items, such as canes and walkers, out of the person’s reach.

Provide information in a way that does not work for your customer.

Shout.

Use obscure and incomprehensible language.

Walk away without saying good-bye.

I think that many of the people who actually need many of the tips given aren’t going to be helped at all by them since this likely indicates a failure to observe their environment.

That said, I think I still did learn some things from the education. One is that most people that are legally blind are not fully blind. I also admit that I don’t understand all of the points. For example, if I encountered somebody with a deafblind impairment who was accompanied by an “Intervenor”, I would guess that I’d address my communication at the Intervenor because I could communicate with that individual. Perhaps that point was meant only for Intervenors for less severe communication issues, such as communication with somebody deaf accompanied by somebody who signs for them?

I’d also guess that the tendency to want to help causes many people to violate the “Ask before you help” point without them even realizing it, but if the individual doesn’t mention it when it happens, being told to ask first is probably not enough. That must be frustrating to an involuntarily “helped” individual.

Chapter 17.

\begin{itemize}
\item Page 297. (17.38). appears to be off by a factor of since .
\item Page 311. (17.134). missing on the term in the integral.
\item Page 311. (17.136). First term (non-integral part) should be negated.
\item Page 312. (17.144). Sign on before sum positive instead of negative.
\item Page 313. (17.149). missing.
\item Page 313. (17.152,17.154). extra bra around the bra.
\item Page 313. (17.153). bra missing on
\end{itemize}

Chapter 24.

\begin{itemize}
\item Page 450. (24.6). should be .
\item Page 452. (24.18). In the case should be instead of (although what's in the text is strictly still correct since it only changes the phase of the wavefunction).

\item Page 455. (24.40). RHD should be multiplied by .

\item Page 460. (24.71). should be .
\item Page 460. Third paragraph. should be .
\item Page 460. (24.76). The integral should be , not . This messes up some of the subsequent stuff, unless there is also another compensating error. Note that one can check this easily since the derivative of is .
\end{itemize}

Chapter 25.

Chapter 26.

\begin{itemize}
\item Page 486. (26.60). ought to have braces and read .
\item Page 487. (26.67). in the position should be .
\item Page 489. before (26.84). For rotations about the imaginary axis was probably meant to be the i’th axis.
\item Page 495. (26.149,26.150). Looks like ‘s are missing (esp. compared to 26.144-145).
\item Page 495. (26.150). off by . () as is.
\item Page 496. (26.154). An extra in the integral, in between and the .
\item Page 498. (26.175). should be in the first line.
\item Page 498. (26.178). An factor has been lost in either (26.178) or (26.179).
\item Page 499. (26.190). minor: should be .
\item Page 450. (26.192). minor: should be , and should be .
\end{itemize}

Chapter 27.

\begin{itemize}
\item Page 503. (27.8). minor: bold . probably meant to be .
\item Page 504. (27.20). Missing factor on LHS.
\item Page 507. before (27.53). minor: Velocity missing bold.
\item Page 510. before (27.78). minor: periods in the two kets should be commas.
\item Page 510. (27.80). and should be interchanged (if is the polar angle then for that rotation, and for the rotation in the plane). The ‘s here should also be dropped.
\item Page 511. (27.81). Same as 27.80.
\item Page 511. (27.83). ‘s should be dropped.
\item Page 514. (27.109). minor: dot instead of cdot.
\item Page 515. (27.117). Same error as in (26.149-150). ‘s missing, and wrong sign on .
\end{itemize}

1D QM scattering. No potential wave packet time evolution.

Now lets move to the QM picture where we assume that we have a particle that can be represented as a wave packet as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig3}
\caption{Wave packet for a particle wavefunction }
\end{figure}

First without any potential , lets consider the evolution. Our position and momentum space representations are related by

and by Fourier transform

Schr\”{o}dinger’s equation takes the form

or more simply in momentum space

Rearranging to integrate we have

and integrating

or

Time evolution in momentum space for the free particle changes only the phase of the wavefunction, the momentum probability density of that particle.

With a potential.

Now “switch on” a potential, still assuming a wave packet representation for the particle. With a positive (repulsive) potential as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig6}), at a time long before the interaction of the wave packet with the potential we can visualize the packet as heading towards the barrier.

After some time long after the interaction, classically for this sort of potential where the particle kinetic energy is less than the barrier “height”, we would have total reflection. In the QM case, we’ve seen before that we will have a reflected and a transmitted portion of the wave packet as depicted in figure (\ref{fig:qmTwoL21:qmTwoL21Fig7})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig7}
\caption{QM wave packet long after interaction with repulsive potential.}
\end{figure}

Even if the particle kinetic energy is greater than the barrier height, as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig8}), we can still have a reflected component.
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig8}
\caption{Kinetic energy greater than potential energy.}
\end{figure}

This is even true for a negative potential as depicted in figure (\ref{fig:qmTwoL21:qmTwoL21Fig9})!

Consider the probability for the particle to be found anywhere long after the interaction, summing over the transmitted and reflected wave functions, we have

Observe that long after the interaction the cross terms in the probabilities will vanish because they are non-overlapping, leaving just the probably densities for the transmitted and reflected probably densities independently.

We define

The objective of most of our scattering problems will be the calculation of these probabilities and the comparisons of their ratios.

Question. Can we have more than one wave packet reflect off. Yes, we could have multiple wave packets for both the reflected and the transmitted portions. For example, if the potential has some internal structure there could be internal reflections before anything emerges on either side and things could get quite messy.

Considering the time independent case temporarily.

We are going to work through something that is going to seem at first to be completely unrelated. We will (eventually) see that this can be applied to this problem, so a bit of patience will be required.

We will be using the time independent Schr\”{o}dinger equation

where we have added a subscript to our wave function with the intention (later) of allowing this to vary. For “future use” we define for

Consider a potential as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig10}), where for and .

We won't have bound states here (repulsive potential). There will be many possible solutions, but we want to look for a solution that is of the form

Suppose , we have

Defining

we write Schr\”{o}dinger’s equation as a pair of coupled first order equations

At this specifically, we “know” both and and have

This allows us to find both

then proceed to numerically calculate and at neighboring points . Essentially, this allows us to numerically integrate backwards from to find the wave function at previous points for any sort of potential.

I had a three thread timing hole scenerio that I wanted to confirm with the debugger. Adding blocks of code to selected points like this turned out to be really handy:

{
volatile int loop = 1 ;
while (loop)
{
loop = 1 ;
sleep(1) ;
}
}

Because the variable loop is local, I could have two different functions paused where I wanted them, and once I break on the sleep line, can let each go with a debugger command like so at exactly the right point in time

(gdb) p loop=0

(assigns a value of zero to the loop variable after switching to the thread of interest). The gdb ‘set scheduler-locking on/off’ and ‘info threads’ ‘thread N’ commands are also very handy for this sort of race condition debugging (this one was actually debugged by code inspection, but I wanted to see it in action to confirm that I had it right).

I suppose that I could have done this with a thread specific breakpoint. I wonder if that’s also possible (probably). I’ll have to try that next time, but hopefully I don’t have to look at race conditions like today’s for a quite a while!

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Rotations of operators.

READING: section 28 [1].

Rotating with as in figure (\ref{fig:qmTwoL19:qmTwoL19fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL19fig1}
\caption{Rotating a state centered at }
\end{figure}

So

Any three operators that transform according to

form the components of a vector operator.

Infinitesimal rotations

Consider infinitesimal rotations, where we can show that

Note that for we recover the familiar commutator rules for angular momentum, but this also holds for operators , , , …

Note that

so

so

In the same way, suppose we have nine operators

that transform according to

then we will call these the components of (Cartesian) a second rank tensor operator. Suppose that we have an operator that transforms

Then we will call a scalar operator.

A problem.

This all looks good, but it is really not satisfactory. There is a problem.

Suppose that we have a Cartesian tensor operator like this, lets look at the quantity

We see buried inside these Cartesian tensors of higher rank there is some simplicity embedded (in this case trace invariance). Who knows what other relationships are also there? We want to work with and extract the buried simplicities, and we will find that the Cartesian way of expressing these tensors is horribly inefficient. What is a representation that doesn’t have any excess information, and is in some sense minimal?

How do we extract these buried simplicities?

Recall

gives a linear combination of the .

We’ve talked about before how these form a representation of the rotation group. These are in fact (not proved here) an irreducible representation.

Look at each element of . These are matrices and will be different according to which rotation is chosen. There is some for which this element is nonzero. There’s no element in this matrix element that is zero for all possible . There are more formal ways to think about this in a group theory context, but this is a physical way to think about this.

Think of these as the basis vectors for some eigenket of .

where

So

where

Recall that

Define operators , as the elements of a spherical tensor of rank if

Here we are looking for a better way to organize things, and it will turn out (not to be proved) that this will be an irreducible way to represent things.

Examples.

We want to work though some examples of spherical tensors, and how they relate to Cartesian tensors. To do this, a motivating story needs to be told.

Let’s suppose that is a ket for a single particle. Perhaps we are talking about an electron without spin, and write

for and after dropping . So

We are writing this in this particular way to make a point. Now also assume that

Disclaimer.

Recap.

First column

Let’s start with computation of the kets in the lowest position of the first column, which we will obtain by successive application of the lowering operator to the state

Recall that our lowering operator was found to be (or defined as)

so that application of the lowering operator gives us

Proceeding iteratively would allow us to finish off this column.

Second column

Moving on to the second column, the top most element in the table

can only be made up of with . There are two possibilities

So for some and to be determined we must have

Observe that these are the same kets that we ended up with by application of the lowering operator on the topmost element of the first column in our table. Since and are orthogonal, we can construct our ket for the top of the second column by just seeking such an orthonormal superposition. Consider for example

With we find that , so we have

So we find, for real and that

for any orthonormal pair of kets and . Using this we find

This will work, although we could also multiply by any phase factor if desired. Such a choice of phase factors is essentially just a convention.

The Clebsch-Gordon convention

This is the convention we will use, where we

\begin{itemize}
\item choose the coefficients to be real.
\item require the coefficient of the term to be
\end{itemize}

This gives us the first state in the second column, and we can proceed to iterate using the lowering operators to get all those values.

Moving on to the third column

can only be made up of with . There are now three possibilities

and 2 orthogonality conditions, plus conventions. This is enough to determine the ket in the third column.

We can formally write

where

and

are the Clebsch-Gordon coefficients, sometimes written as

Properties
\begin{enumerate}
\item only if

This is sometimes called the triangle inequality, depicted in figure (\ref{fig:qmTwoL18:qmTwoL18fig1})

rotates in the same way. Rotating a ket as in figure (\ref{fig:qmTwoL18:qmTwoL18fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL18fig3}
\caption{Rotating a wavefunction.}
\end{figure}